Welcome to MathOverflow

Question

8

3

Does there exist a continuous function $f:[0,1]\rightarrow [0,1]$ such that $f$ takes every value in $[0,1]$ an infinite number of times?

flag	asked Mar 18 2010 at 21:59 Cristos A. Ruiz 661●6●10

Tom Leinster · Accepted Answer · 2010-03-18 22:04:24Z UTC

23

Yes. In fact, there exists such an $f$ taking every value uncountably many times.

Take a continuous surjection $g: [0, 1] \to [0, 1]^2$. (Such things exist: they're space-filling curves.) Then the composite $f$ of $g$ with first projection $[0, 1]^2 \to [0, 1]$ has the required property.

answered Mar 18 2010 at 22:04

Tom Leinster
13.5k●1●40●82

2

Great I didn't thought of that one, can I ask differentiability? – Cristos A. Ruiz Mar 18 2010 at 22:14

6

Yes and en.wikipedia.org/wiki/Volterra's_function is an example. – Johannes Hahn Mar 18 2010 at 22:20

Is it obvious that these curve are continuous? – Tran Chieu Minh Mar 19 2010 at 7:52

1

Space-filling curves, such as the Peano curve, are indeed continuous. The existence a not-necessarily-continuous surjection from [0, 1] to [0, 1]^2 is much easier to prove --- it's just an argument about cardinality of sets. Often one defines a space-filling curve as a uniform limit of a sequence of curves, and uses the fact that a uniform limit of continuous maps is continuous. – Tom Leinster Mar 19 2010 at 14:03

Petya · Answer 2 · 2010-03-18 22:04:35Z UTC

14

Take a projection of a well-known Peano curve, which is a surjective continuous mapping $[0,1] \to [0,1]^2$ to a factor.

answered Mar 18 2010 at 22:04

Petya
2,955●6●24

1

Snap ! – Tom Leinster Mar 18 2010 at 22:07

You are few seconds faster, Tom. – Petya Mar 18 2010 at 22:08

Continuous function from $[0,1]$ to $[0,1]$

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.